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Geometry problem(Again a Challenge!!)

In a triangle \(ABC\),\(\angle BAC = 60^{\circ},AB=2AC.\)Point P is inside the triangle such that \(PA=\sqrt{3}, PB=5,PC=2.\) What is the area of triangle \(ABC\)?

This is for those who are missing geometry section.The solution which I know is extremely surprising and beautiful but I want an alternate solution because the idea of solution which I know is impossible for a normal student to strike in his mind, even might be tough for good students.I think it might be easy for brilliant users who are good at geometry.So I ask all of you for an alternate solution.I think there are solutions using trigonometry but they are very ugly involving big big equations and variables.Let's see who solve it first.!It might be annoying because I am starting many discussions nowadays,its because RMO is coming nearer.Thanks in advance!

then
is a right angle triangle with sides b;a and 2b so
WE NAME
so
we again use cosine rule-
and
or
now
or
simflifying it will get
putting
from this equation we will get the value of b now the area of the triangle
so putting the values of b we will get the area of ABC.
by sridhar acharya's theorem

see that if we apply cosine rule - then we have
\[a=\sqrt{3}b\] then \[\Delta ABC\] is a right angle triangle with sides b;a and 2b
so \[\angle ACB=90\]
WE NAME \[\angle ACP=x\] so \[\angle BCP=90-x\]
we again use cosine rule-
\[cosx=\frac{1+b^{2}}{4b}\] and \[cos(90-x)=\frac{a^{2}-21}{4a}\] or \[sinx=\frac{a^{2}-21}{4a}\]
now \[(sinx)^{2}+(cosx)^{2}=1\] or \[\frac{(a^{2}-21)^{2}}{16a^{2}}+\frac{(1+b^{2})^2}{16b^{2}}=1\]
simflifying it will get \[b^{4}-14b^{2}+37=0\] putting \[a=\sqrt{3}b\]
from this equation we will get the value of b
now the area of the triangle \[X=\frac{1}{2}bcsin60=\frac{\sqrt{3}}{2}b^{2}\]
so putting the values of b we will get the area of ABC. \[b=7+2\sqrt{3}\] by sridhar acharya's theorem

We know the triangle \(ABC\) is a 30-60-90 right triangle. We can use coordinates. Let \(a>0, C(0,0), A(a,0), B(0, \sqrt{3}a), P(x,y).\) Using distance formula, we have 3 equations. It's easy to deduce that \(a^4-14a^2+37=0.\) Hence \(a^2=7 \pm 2\sqrt{3}.\) If \(a^2=7 - 2\sqrt{3}.\), then 2 is the longest side of the triangle \(PAC\). Hence \(\angle{PAC} > 60^{\circ}.\) Therefore the point P is outside of triangle \(ABC\). So \(a^2=7 + 2\sqrt{3}\), and the area is simply \(\frac{\sqrt{3}}{2}a^2 = \frac{7\sqrt{3}}{2} + 3.\). Notice that \( \angle{APC} = 120^{\circ}\) by Law of Cosines, some creative mind may construct a pure geometric argument from this info.

Let \(X, Y, Z\) be the reflections of point \(P\) w.r.t the sides \(CA, AB, BC\), then it's easy to show the triangle \(XYZ\) is a 3-4-5 triangle. So it's a right triangle with \(\angle ZXY=90^{\circ}\). Hence \(AC^2=7+2\sqrt{3}\) since \(CX=2, XA=\sqrt{3}\) and \(\angle CXA = 120^{\circ}.\)

Or, we can add areas of three triangles \(BYZ\) (an equilateral with sides=5), \(XYZ\) (3-4-5 right triangle), and \(XYA\) (120 degree isoceles triangle with sides \(\sqrt{3},\sqrt{3},3\) ) together which is two times the area of the triangle \(ABC\).

Yeah..this was the basic idea behind the problem.One needs to reflect the point P about all the sides and then there will be a pentagon(not hexagon) forming with these vertices.Find the area of the pentagon in 2 different ways and you will get your answer without trigonometry or coordinate geometry. By this method,the solution will be short and very beautiful because it involves extreme creativity.!!

The triangle is a 30-60-90 triangle since one angle is 60 which is give and one side is twice another side (precisely the side opposite to 30 must be half the hypotenuse). If we assign the value \(x\) to the hypotenuse the side opposite to 30 must be \(x/2\) and that opposite to 60 must be \(3^.5*x/2\). Then we can get an equation in \(x\) using Heron's formula for the areas of the three smaller triangles the sum of which must be equal to the base \(*\) height of the 30-60-90 triangle. Solving this equation is a whole new ball game though . I doubt you could do it without Wolfram.